Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 MRRProof (⇔)
↳8 QDP
↳9 QDPOrderProof (⇔)
↳10 QDP
↳11 PisEmptyProof (⇔)
↳12 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a(a(y, 0), 0) → y
c(c(y)) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a(x1, x2)) = x1 + x2   
POL(c(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

c(c(y)) → y


(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

C(a(c(c(y)), x)) → A(c(c(c(a(x, 0)))), y)
C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
C(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → C(a(x, 0))
C(a(c(c(y)), x)) → A(x, 0)

The TRS R consists of the following rules:

a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

C(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
C(a(c(c(y)), x)) → C(a(x, 0))

The TRS R consists of the following rules:

a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

C(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → C(a(x, 0))


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(C(x1)) = x1   
POL(a(x1, x2)) = x1 + x2   
POL(c(x1)) = 1 + x1   

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

C(a(c(c(y)), x)) → C(c(c(a(x, 0))))

The TRS R consists of the following rules:

a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(C(x1)) = -I + 1A·x1

POL(a(x1, x2)) = -I + 1A·x1 + 4A·x2

POL(c(x1)) = 5A + 1A·x1

POL(0) = 0A

The following usable rules [FROCOS05] were oriented:

a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)

(10) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(12) TRUE